Fix typos

Signed-off-by: Marcello Seri <[email protected]>

1-manifolds.tex

@@ -186,7 +186,7 @@ \section{Differentiable manifolds}
 \begin{enumerate}[(i)]
   \item\label{thm:chainrule1} The function $g\circ f: U\subset\R^n \to\R^m$ is continuously differentiable and its total derivative~\eqref{eq:jacobian} at a point $x\in U$ is given by
 \begin{equation}
-  D(g\circ f)(x) = D(g(f(x)) \circ Df(x).
+  D(g\circ f)(x) = (Dg)(f(x)) \circ Df(x).
 \end{equation}
 \item\label{thm:chainrule2} Denote $x=(x^1, \ldots, x^n)\in\R^n$ and $y=(y^1,\ldots,y^k)\in\R^k$ the coordinates on the respective euclidean spaces and $f=(f^1,\ldots,f^k)$ and $g=(g^1,\ldots,g^m)$ the components of the functions. Then the partial derivatives of $g\circ f$ are given by
 \begin{equation}
@@ -503,7 +503,7 @@ \section{Differentiable manifolds}
 
   \begin{marginfigure}
     \includegraphics{1_2_25-sphere}
-    \caption{The identification $\sim$ of antipodal points maps the sphere to a disk. Embedding $\bS^n/\!\sim$ in $\R^{n+1}$, one can define a map $\pi_D$ that projects the representative of $[x]$ in the north hemisphere orthogonally to the disk $D^n = \{x\in\R^{n+1} \mid \|x\|\leq 1, \; x^{n+1}=0\}$ (the equator is mapped to itself). }
+    \caption{The identification $\sim'$ of antipodal points maps the sphere to a disk. Embedding $\bS^n/\!\sim'$ in $\R^{n+1}$, one can define a map $\pi_D$ that projects the representative of $[x]$ in the north hemisphere orthogonally to the disk $D^n = \{x\in\R^{n+1} \mid \|x\|\leq 1, \; x^{n+1}=0\}$ (the equator is mapped to itself). }
   \end{marginfigure}
   There is a nice interpretation of this construction in terms of flattening spheres.
   Observe that a line through the origin always intercepts a sphere $\bS^n$ at two antipodal points and, conversely, each pair of antipodal point determines a unique line through the center.
@@ -512,7 +512,7 @@ \section{Differentiable manifolds}
   Note that by gluing antipodal points, we are identifying the north and south hemispheres, thus essentially flattening the sphere to a disk.
 
   \begin{exercise}\label{exe:RPSN}
-    Show that the map $n: \R^{n+1}_0\to \bS^n$, $n(x) = \frac{x}{\|x\|}$, induces a homeomorphism $\hat n:\RP^n \to \bS^n/\!\sim$.\\
+    Show that the map $n: \R^{n+1}_0\to \bS^n$, $n(x) = \frac{x}{\|x\|}$, induces a homeomorphism $\hat n:\RP^n \to \bS^n/\!\sim'$.\\
     \textit{\small Hint: find an inverse map and show that both $\hat n$ and its inverse are continuous.}
   \end{exercise}
 
@@ -669,7 +669,7 @@ \section{Smooth maps and differentiability}
 \begin{definition}
 A \emph{diffeomorphism} $F$ between two smooth manifolds $M_1$ and $M_2$ is a bijective map such that $F\in C^\infty(M_1, M_2)$ and $F^{-1}\in C^\infty(M_2, M_1)$.
 %
-Two smooth manifolds $M_1$ and $M_2$ are called \emph{diffeomporphic} if there exists a diffeomorphism $F:M_1\to M_2$ between them.
+Two smooth manifolds $M_1$ and $M_2$ are called \emph{diffeomorphic} if there exists a diffeomorphism $F:M_1\to M_2$ between them.
 \end{definition}
 
 \begin{exercise}
@@ -838,7 +838,7 @@ \section{Partitions of unity}
 
 \section{Manifolds with boundary}\label{sec:mbnd}
 
-\newthought{The definition of manifolds has a serious limitation}, even though it is perfectly good to describe curves\footnote{E.g. the circle seen in Example~\ref{ex:S1emb}.} and surfaces\footnote{E.g. the $n
+\newthought{The definition of manifolds has a serious limitation}, even though it is perfectly good to describe curves\footnote{E.g. the circle seen in Example~\ref{ex:S1emb}.} and surfaces\footnote{E.g. the $
 2$-spheres $\bS^2$ from the homework sheet.}, it fails to describe many natural objects like a \emph{closed} interval $[a,b]\in\R$ or the \emph{closed} disk $D_1(0)$ of Example~\ref{ex:uball}.
 Note that in each of these cases, both the interior and the boundary are smooth manifolds and their dimension differ by one\footnote{In the first case the interior $(a,b)$ is a $1$-manifold and the boundary, the set $\partial[a,b] = \{a,b\}$, is a $0$-manifold. In the second case the interior of $D_1(0)$ is the open unit ball, a $2$-manifold, and the boundary $\partial D_1(0)$ is the $1$-manifold $\bS^1$.}.
 
@@ -910,7 +910,7 @@ \section{Manifolds with boundary}\label{sec:mbnd}
 \end{definition}
 
 \begin{proposition}\label{prop:bdwelldef}
-  The boundary $\partial M$ is well--defined\footnote{In the sense that smooth charts send boundary pieces to boundary pieces.Note that the definition of the boundary holds for topological manifolds as well, but showing that it is well--defined is much more complicated and will be omitted.}.
+  The boundary $\partial M$ is well--defined\footnote{In the sense that smooth charts send boundary pieces to boundary pieces. Note that the definition of the boundary holds for topological manifolds as well, but showing that it is well--defined is much more complicated and will be omitted.}.
 \end{proposition}
 \begin{proof}
   The statement follows if we show that the transition maps send boundary pieces to boundary pieces.
@@ -920,7 +920,7 @@ \section{Manifolds with boundary}\label{sec:mbnd}
   \begin{equation}
     f(x+h) = f(x) + Df|_x h + O(\|h\|).
   \end{equation}
-  Since the total derivative $D f$ at $x$ is an isomorphism, there exist an open neighbourhood $\cO$ of $x$ such that $f(\cO)$ is open in $\R^n$ and thus $f(x)\in V\cap(\cH^n\cap\partial\cH^n)$.
+  Since the total derivative $D f$ at $x$ is an isomorphism, there exist an open neighbourhood $\cO$ of $x$ such that $f(\cO)$ is open in $\R^n$ and thus $f(x)\in V\cap(\cH^n\setminus\partial\cH^n)$.
 \end{proof}
 
 \begin{example}\label{ex:mfldbdryinterval}

2-tangentbdl.tex

@@ -647,7 +647,7 @@ \section{Tangent vectors as tangents to curves}
   \end{equation}
 \end{enumerate}
 
-Do these definition agree?
+Do these definitions agree?
 One way to check is to pick a chart $\varphi: U \to \varphi(U)$ in a neighbourhood of $\gamma(t)$, and compare the expressions in local coordinates. Let $(x^i)$ denote the coordinates of $\varphi$ and define the curves $\gamma^i := x^i \circ \gamma : I\to\R$.
 Let's focus on~\eqref{eq:tg_curve_der}. By definition, $\gamma'(t)(x^i) = (x^i\circ\gamma)'(t) = (\gamma^i)'(t)$, therefore by Proposition~\ref{prop:basis_TpM} we get
 \begin{equation}\label{eq:tg_curve_vec}
@@ -889,7 +889,7 @@ \section{Vector bundles}\label{sec:vectorbundle}
 
 To compare vector bundles it is useful to define the following concept.
 \begin{definition}
-  An \emph{isomorphim} between two vector bundles $\pi_i: E_i \to M$, $i=1,2$, over the same base space $M$ is a homeomorphism $h:E_1 \to E_2$ which maps every fiber $\pi_1^{-1}(p)$ to the corresponding fiber $\pi_2^{-1}(p)$ by a linear isomorphism.
+  An \emph{isomorphism} between two vector bundles $\pi_i: E_i \to M$, $i=1,2$, over the same base space $M$ is a homeomorphism $h:E_1 \to E_2$ which maps every fiber $\pi_1^{-1}(p)$ to the corresponding fiber $\pi_2^{-1}(p)$ by a linear isomorphism.
 \end{definition}
 
 Since an isomorphism preserves all the structure of a vector bundle, isomorphic bundles are often regarded as the same.

3-vectorfields.tex

@@ -48,7 +48,7 @@ \section{Vector fields}
 \end{equation}
 The zero element of the vector space, called \emph{zero vector field}, is the vector field whose values is $0\in T_pM$ for all $p\in M$.
 Moreover, each vector field can be multiplied by smooth functions $f\in C^\infty(M)$ by defining $fX:M\to  TM$ by
-\marginnote{Be carefur, we are talking about two different structures here: $\fX(M)$ is both a real vector space and a $C^\infty(M)$-module.}
+\marginnote{Be careful, we are talking about two different structures here: $\fX(M)$ is both a real vector space and a $C^\infty(M)$-module.}
 \begin{equation}
   (fX)_p = f(p)X_p.
 \end{equation}
@@ -673,7 +673,7 @@ \section{Flows and integral curves}
     &\frac{d}{d t} e^{tX} (p) = X(e^{tX}(p)), \quad \forall p\in M.
   \end{align}
 
-  Moreover, if $X(x) = Ax$ is a linear vector field on $\R^n$, i.e., $A$ is a $n\times n$-matrix, then the corresponding flow $\varphi_t$ is the matrix exponential $\varphi_t(x) = e^{Ax}$.
+  Moreover, if $X(x) = Ax$ is a linear vector field on $\R^n$, i.e., $A$ is a $n\times n$-matrix, then the corresponding flow $\varphi_t$ is the matrix exponential $\varphi_t(x) = e^{tA}x$.
 \end{exercise}
 
 There are a few cases in which we can guarantee completeness, let's have a brief look.
@@ -716,7 +716,7 @@ \section{Flows and integral curves}
 In the same fashion as Lemma~\ref{lemma:charactCompleteVF}, one can characterize non-complete vector fields.
 
 \begin{lemma}\label{lemma:charactNotCompleteVF}
-  Let $K\subset \circ M$ compact and $p\in K$.
+  Let $K\subset \mathring M$ compact and $p\in K$.
   If $t^+(p) < \infty$, then there exists $0<\tau<t^+(p)$ such that $\gamma_p(t)\not\in K$ for all $t\in (\tau, t^+(p))$.
   An analogous statement holds if $t^-(p) > -\infty$.
 \end{lemma}

3b-liegroups.tex

@@ -43,7 +43,7 @@ \section{Lie groups}
 \end{definition}
 
 \begin{example}
-  It turns out that you know plenty of examples of Lie group homomorpisms.
+  It turns out that you know plenty of examples of Lie group homomorphisms.
   \begin{enumerate}
     \item The map $\exp:\R\to\R_+$ is a Lie group homomorphism. The image of $\exp$ is the open subgroup $\R_+$ and $\exp:\R\to\R_+$ is a Lie group isomorphism with inverse $\log:\R_+\to\R$.
     \item The map $\epsilon:\R\to\bS^1$ defined by $\epsilon(t)=e^{2\pi i t}$ is a Lie group homomorphism whose kernel is $\Z$. Similarly, the map $\epsilon^n:\R^n\to\bT^n$ defined by $\epsilon^n(x^1, \ldots, x^n)=(e^{2\pi i x^1}, \ldots, e^{2\pi i x^n})$ is a Lie group homomorphism whose kernel is $\Z^n$.

4-cotangentbdl.tex

@@ -3,7 +3,7 @@ \section{The cotangent space}
 \newthought{The dual of a vector space} should be a well-known concept from linear algebra. We recall it here just for the sake of convenience.
 
 \begin{definition}
-  Let $V$ a vector space of dimension $n\in N$.
+  Let $V$ a vector space of dimension $n\in \N$.
   Its \emph{dual space} $V^* := \cL(V, \R)$ is the $n$-dimensional real vector space of linear maps $\omega:V \to R$.
   The elements of $V^*$ are usually called \emph{linear functionals} and for $\omega\in V^*$ and $v\in V$ it is common to write
   \begin{equation}

6-differentiaforms.tex

@@ -333,7 +333,7 @@ \section{Differential forms on manifolds}
   Let $(x^i)$ and $(y^i)$ are two different local coordinates on some open $V\subset M$.
   Show that the following identity holds:
   \begin{equation}
-    dy^1\wedge dy^n = \det\left(\frac{\partial y^j}{\partial x^i}\right) dx^1\wedge dx^n.
+    dy^1\wedge\cdots\wedge dy^n = \det\left(\frac{\partial y^j}{\partial x^i}\right) dx^1\wedge\cdots\wedge dx^n.
   \end{equation}
 \end{exercise}
 

aom.tex

@@ -208,7 +208,7 @@
     \setlength{\parskip}{\baselineskip}
     Copyright \copyright\ \the\year\ \thanklessauthor
 
-    \par Version 0.9.7 -- \today
+    \par Version 0.9.8 -- \today
 
     \vfill
     \small{\doclicenseThis}
@@ -267,7 +267,7 @@ \chapter*{Introduction}
 In some sense I would like this course to provide the introduction to geometric analysis that I wish was there when I prepared my \href{https://www.mseri.me/lecture-notes-hamiltonian-mechanics/}{first edition} of the Hamiltonian mechanics course.
 
 I am extremely grateful to Martijn Kluitenberg for his careful reading of the notes and his useful comments and corrections.
-Many thanks also to Huub Bouwkamp, Bram Brongers, Mollie Jagoe Brown, Nicol\'as Moro, Luuk de Ridder, Lisanne Sibma, Jordan van Ekelenburg, Hanneke van Harten and Dave Verweg for reporting a number of misprints and corrections.
+Many thanks also to Huub Bouwkamp, Bram Brongers, Mollie Jagoe Brown, Wietze Koops, Nicol\'as Moro, Luuk de Ridder, Lisanne Sibma, Jordan van Ekelenburg, Hanneke van Harten and Dave Verweg for reporting a number of misprints and corrections.
 
 \mainmatter